Recent content by operationsres

Suppose that \sigma(t,T) is a deterministic process, where t varies and T is a constant. We also have that t \in [0,T]. Also W(t) is a Wiener process. My First Question What is \displaystyle \ \ d\int_0^t \sigma(u,T)dW(u)? My lecture slides assert that it's equal to \sigma(t,T)dW(t) but I'm...

Hi, I'm not sure what "modulo" means? It would make sense if the solution to QUESTION1 is actually -f(x), is that what you were saying?

Let f(x) be a non-stochastic mapping f: \mathbb{R} \to \mathbb{R}. The second fundamental theorem of calculus states that: \frac{d}{dx} \int_a^x f(s)ds = f(x). *QUESTION 1* Is the following true? \frac{d}{dx} \int_x^a f(s)ds = f(x). *QUESTION 2* Related to this, how can I...

I don't really understand what's meant when we integrate with respect to a Wiener process. I was never taught this.I know all of the properties of B(t) though. My qualitative, uneducated guess as to what that integral is doing is adding up all the instantaneous changes in B(t) over the...

So your answer is "not sure"? :p

nice, I understand, well done Sammy!

Dear Mute, Since you know about the OU process, can I ask if this is an accurate representation of a mean-reverting OU? dX(t) = (m-X(t))dt + \sigma X(t) dB(t) where m is the mean-reversion term, B(t) is standard Brownian Motion. I ask because (i) This is what's in my tutorial...

In fact it is. Nice!

\displaystyle \ \ \frac{d}{ds}\left(e^{-us} X(s)\right)\ = \int_0^t \frac{d}{ds}\left(e^{-us} X(s)\right)ds \displaystyle \ \ = d(e^{-ut}X(t)) It seems I've gone in a circle (obviously I didn't do what you were asking for).

Working with this, we have that: \int_{0}^{t}\left(\frac{d}{ds}\left(e^{-us} X(s)\right)\right)ds = \int_{0}^{t} \left( -ue^{-us}X(s) + e^{-us} \right)ds = -u \int_0^t e^{-us}X(s)ds + \int_0^t e^{-us}ds So I'm still not sure how I can get to the identity based on what you've provided?

We all know that \int_0^t dB(s) = B(t), where B(t) is a standard Brownian Motion. However, is this identity true? \int_{t_1}^{t_2} dB(s) = B(t_2) - B(t_1)

Why does: \int_0^t d(e^{-us} X(s)) = \sigma \int_0^t e^{-us} dB(s) for stochastic process X(t) and Wiener process B(t)? Also, why is the following true: \int_0^t d(e^{-us} X(s)) = e^{-ut}X(t) - X(0)

Thanks, 1) What attached material? 2) I'm at the end of a financial mathematics course (stochastic calculus). Integrating factors are provided to us and we will never learn how to discover them. I want to learn how to do this -- they aren't going to teach this to me.

Whenever I'm given a SDE problem that requires us to multiply both sides by an "integrating-factor", it's always given to us as a *Hint*. I would like to know how to come up with these integrating factors. Here's some examples: 1) For the mean-reverting Ornstein-Uhlenbeck (OU) SDE dX_t =...

Suppose that B(t) is a Wiener process. Suppose that the following equation is true: B(t)(t + \frac13 B(t)) = B(t)^{0.5}. I've conjured this equation out of thin air (it's probably not true) to ask the following question. Does the above identity (assuming it's correct) enable us to write the...